Answer
$x=\left\{- \frac{1}{2}, 5 \right\}$
Work Step by Step
The factor theorem says that if $f(c)=0$, then $(x-c)$ is a factor of $f(x)$, and if $(x-c)$ is a factor of $f(x)$, then $f(c)=0$.
We check $x=-1$:
$f(-1)=2(-1)^4-13(-1)^3+7(-1)^2+37(-1)+15=0$ and
We check $x=3$:
$f(3)=2(3)^4-13(3)^3+7(3)^2+37(3)+15=0$
thus, it is indeed a zero.
Then, we can write:
$f(x)=(x+1)(x-3)(2x^2-9x-5)$.
Factorize the trinomial factor $(2x^2-9x-5)$
(find two factors of $2(-5)=-10$ whose sum is $-9$):
($-10$ and $+1$).
$2x^2-9x-5=2x^2+x-10x-5$ ...factor in pairs...
$=x(2x+1)-5(2x+1)=(2x+1)(x-5)$.
$f(x)=(x+1)(x-3)(2x+1)(x-5)$.
Thus the other zeros are:
$$x=\left\{- \frac{1}{2}, 5 \right\}$$
